Response to 6.3.51
For this problem, we could take both sqrt(sin^3 t * cos t) and sqrt(cos t) and combine them to get sqrt(sin^3 t * cos^2 t) because it is ok to multiply square root values.
Next, we can actually calulate the square root of this eq., and the answer comes to be
sqrt(sin^2 t * sin t * cos^2 t) = abs(sin t * cos t) * sqrt(sin t)
We need absolute value signs b/c the values of sqrt's can't be negative. As far as geting to the answer, it is a matter of breaking up the multiples.
Next, we can actually calulate the square root of this eq., and the answer comes to be
sqrt(sin^2 t * sin t * cos^2 t) = abs(sin t * cos t) * sqrt(sin t)
We need absolute value signs b/c the values of sqrt's can't be negative. As far as geting to the answer, it is a matter of breaking up the multiples.
posted by battlejoe18 at 9:13 PM
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